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Home » Thus, the lateral surface area of the cone is \(\boxed60\pi\). - American Beagle Club

Thus, the lateral surface area of the cone is \(\boxed60\pi\). - American Beagle Club

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 -4y^2 + 12y + 1 \geq 0

📌 Multiply through by \(-1\) (reversing the inequality):

📌 4y^2 - 12y - 1 \leq 0

📌 Solve the quadratic inequality using the quadratic formula:

📌 y = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{12 \pm \sqrt{144 + 16}}{8} = \frac{12 \pm \sqrt{160}}{8}

📌 y = \frac{12 \pm 4\sqrt{10}}{8} = \frac{3 \pm \sqrt{10}}{2}

📌 Thus, the range of \(f(x)\) is:

📌 \boxed{\left

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